Lake Preston's latitude is about 44.4 degrees north of the equator. Which global wind belt are

N

9966 [12]

Answer: de?

Explanation:

7 0

3 years ago

Sedimentary layers that are deposited on an angle are called?

Akimi4 [234]

Sedimentary rocks are deposited in layers as strata, forming a structure called bedding.

<span>Bedding planes are surfaces that separate one layer from another. Bedding planes can also form when the upper part of a sediment layer is eroded away before the next episode of deposition. Strata separated by a bedding plane may have different grain sizes, grain compositions, or colors. Sometimes these other traits are better indicators of stratification as bedding planes may be very subtle.</span>

8 0

3 years ago

The radiation per unit area from the Sun reaching the earth is 1400 W/m2 , approximately the amount of radiative power per unit

natka813 [3]

Answer:

83.2 W/m^2

Explanation:

The radiation per unit area of a star is directly proportional to the power emitted, which is given by Stefan-Boltzmann law:

$P=\sigma A T^4$

where

$\sigma$ is the Stefan-Boltzmann constant

A is the surface area

T is the surface temperature

So, we see that the radiation per unit area is proportional to the fourth power of the temperature:

$I \propto T^4$

So in our problem we can write:

$I_1 : T_1^4 = I_2 : T_2^4$

where

$I_1 = 1400 W/m^2$ is the power per unit area of the present sun

$T_1 = 5800 K$ is the temperature of the sun

$I_2$ is the power per unit area of sun X

$T_2 = 2864 K$ is the temperature of sun X

Solving for I2, we find

$I_2 = \frac{I_1 T_2^4}{T_1^4}=\frac{(1400 W/m^2)(2864 K)^4}{(5800 K)^4}=83.2 W/m^2$

6 0

3 years ago

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

nekit [7.7K]

Given:

density of air at inlet, $\rho_{a} = 1.20 kg/m_{3}$

density of air at inlet, $\rho_{b} = 1.05 kg/m_{3}$

Solution:

Now,

$\dot{m} = \dot{m_{a}} = \dot{m_{b}}$

$\rho_{a} A v_{a} = \rho _{b} Av_{b}$ (1)

where

A = Area of cross section

$v_{a}$ = velocity of air at inlet

$v_{b}$ = velocity of air at outlet

Now, using eqn (1), we get:

$\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}$

$\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05}$ = 1.14

% increase in velocity = $1.14\times 100$ =114%

which is 14% more

Therefore % increase in velocity is 14%

5 0

3 years ago

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A reconnaissance plane flies 404 km awayfrom its base at 730 m/s, then flies back to its base at 1095 m/s.What is it’s average s

elena-14-01-66 [18.8K]

The average speed of the plane is 875.999 m/s.

Average speed can be defined as the ratio of total distanced travelled by the object to that of total time taken to cover the distance.

Mathematically, Average speed = Av = $\frac{Total Distance}{Total Time}$

According to the question,

Speed of the plane away from its base V₁ = 730 m/s

Speed of the plane when it flies back V₂ = 1095 m/s

Plane flies the distance D = 404 km

Total Distance covered by the plane S = 404 * 2 km

(because the distance travelled by the plane when going away from the base and then flying back to the base is same)

Therefore S = 808 km = 808 ˣ 10³ m

Time taken by the plane while flying away from the base T₁ = $\frac{D}{V1}$

T₁ = $\frac{404000}{730}$ = 553.425 s

Time taken by the plane while flying back to the base T₂ = $\frac{D}{V2}$

T₂ = $\frac{404000}{1095}$ = 368.949s

Total Time T = T₁ + T₂ = 922.375 s

Therefore Av = $\frac{Total Distance}{Total Time}$

= $\frac{D}{T}$ m/s

= $\frac{808000}{922.375}$ m/s

= 875.999 m/s

The average speed of the plane will be 875.999 m/s.

To know more about Average speed,

brainly.com/question/28641761

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7 0

1 year ago